complex-valued continuous function on $\mathbb{D}$. Would anyone mind providing a hint for the following exercise: Assume $f$ is continuous on the unit disk $\mathbb{D}$  and $\text{Re}(\overline{z}f(z)) > 0$ for all $|z| = 1$. Show that $f(z) = 0$ for some $z$ in the disk. 
I attempted using convexity of the disk, but this didn't lead me very far. 
 A: The Argument provided by Hellen in the case of analytic functions should also work for continuous functions.
Lets assume that $f(z)\neq0$ for all $z$ with $|z|\leq 1$.
If we divide $f(z)$ by $|f(z)|$, the requirement still hold.
Therefore, we can assume that $|f(z)|=1$ ($\forall z\in\mathbb D$) in the original question.
Now consider the closed paths $\gamma_r$ that circles of radius $r$ and center $0$.
The condition $\Re(\bar z f(z))>0$ implies that the image of $f$ under $\gamma_1$
winds around the origin exactly once.
By continuity, the image of $\gamma_r$ for $r\in (0,1)$ also wind around the origin
exactly once.
However, for $r\to0$ this is not possible, because $f(0)$ has to be a single point.
Thus, this yields a contradiction to the original assumption that $f(z)\neq0$.
I suspect that there might be a proof that is more beautiful (and does not rely on arguments that make use of paths, etc.), maybe someone else can find such a proof here.
A: We shall use the following two:
Claim. It suffices to prove the result in the case when $f$ is sufficiently smooth.
Lemma. If $\gamma: [0,2\pi]\to\mathbb C$ is a smooth closed curve, such that $\gamma(t)\ne 0$, for all $z\in [0,2\pi]$, then
$$
\frac{1}{2\pi i}\int_0^{2\pi}\frac{\gamma'(t)\,dt}{\gamma(t)}\in\mathbb Z.
$$
Assume the above hold, and hence that $f$ is smooth, and that $f(z)\ne 0$, for all $z\in\mathbb D$, we set
$$
\gamma(t)=\frac{f(\mathrm{e}^{it})}{|\,f(\mathrm{e}^{it})|}, \quad
\text{for all}\,\,\, t\in[0,2\pi].
$$
Clearly, $\gamma$ is a non-vanishing smooth curve. Also,
$$
\gamma_\lambda(t)=\lambda \gamma(t)+(1-\lambda)\mathrm{e}^{it}\ne 0,
\quad
\text{for all}\,\,\, t\in[0,2\pi],
$$ 
since, as a consequence of the assumption $\mathrm{Re}(\,\overline{z}\,f(z)>0$, we have
$$
\mathrm{Re}\big(\mathrm{e}^{-it}\gamma_\lambda(t)\big)=\mathrm{Re}\left(\lambda \mathrm{e}^{-it}\frac{f(\mathrm{e}^{it})}{|\,f(\mathrm{e}^{it})|}+(1-\lambda)\right)=\frac{\lambda}{|\,f(\mathrm{e}^{it})|}\mathrm{Re}\big(\mathrm{e}^{-it}f(\mathrm{e}^{it})\big)+(1-\lambda)>0.
$$
We next set $\displaystyle H(\lambda)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{\gamma'_\lambda(t)\,dt}{\gamma_\lambda(t)}$. Clearly, $H(\lambda)$ is continuous for $\lambda\in [0,1]$ and its values are in $\mathbb Z$, and hence it is constant. In particular,
$$
\frac{1}{2\pi i}\int_0^{2\pi}\frac{\gamma'(t)\,dt}{\gamma(t)}=H(1)=H(0)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{i\,\mathrm{e}^{it}\,dt}{\mathrm{e}^{it}}=1.
$$
Finally, we set $\zeta_r(t)=\displaystyle \frac{f(r\mathrm{e}^{it})}{|\,f(r\mathrm{e}^{it})|}$ and $\displaystyle G(r)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{\zeta'_r(t)\,dt}{\zeta_r(t)}$, $r\in [0,1]$. Once again,
$G(r)$ is a continuous function of $r$, with values in $\mathbb Z$, and hence it is constant, and in particular
$$
1=G(1)=G(0)=0.
$$ 
We were led to this contraction, having assumed that $f$ does not vanish.
Proof of the Claim. The continuous function $f$ can be extended continuously to, say $K=[-2,2]\times[-2,2]$, and there its extension can be approximated uniformly by polynomials in $x$ and $y$. So if $f_n\to f$ uniformly in the closed disk, and $f_n(z_n)=0$, for some some $z_n$ in the disk, then there exist a converging subsequence $z_{n_k}\to z\in \mathbb D$. Uniform convergence of the $f_n$'s guarantees that $0=f_{n_k}(z_{n_k})\to f(z)$. Clearly, $z$ lies in the interior of the disk, since $f$ does not vanish on the unit circle.
Proof of the Lemma. Set 
$$
g(t)=\gamma(t)\exp\left(-\int_0^t\frac{\gamma'(s)\,ds}{\gamma(s)}\right).
$$
Clearly, $g'(t)=0$, and hence
$$
\gamma(0)=g(0)=g(2\pi)=\gamma(2\pi)\exp\left(-\int_0^{2\pi}\frac{\gamma'(s)\,ds}{\gamma(s)}\right)=\gamma(0)\exp\left(-\int_0^{2\pi}\frac{\gamma'(s)\,ds}{\gamma(s)}\right).
$$
Thus $\displaystyle\exp\left(-\int_0^{2\pi}\frac{\gamma'(s)\,ds}{\gamma(s)}\right)=1$, and hence $\displaystyle\int_0^{2\pi}\frac{\gamma'(s)\,ds}{\gamma(s)}=2k\pi i$, for some $k\in\mathbb Z$.
